Question: s involving definite integrals (algebraic) AP.CALC: CHA‑4 (EU), CHA‑4.D (LO), CHA‑4.D.1 (EK), CHA‑4.D.2 (EK), CHA‑4.E (LO), CHA‑4.E.1 (EK) Google Classroom Facebook Twitter Email You might need: Calculator Problem Consumer surplus is the amount of money that consumers save because they are able to buy a product at a price lower than the highest price they would be willing to pay. The consumer surplus for a certain product increases at a rate of $\dfrac{900}{(x+13)}-35$ dollars per thousand units of the product sold (where $x$ is the number of thousands of units sold). By how many dollars does the surplus increase between $x=7$ and $x=12$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $900\ln(0.8)-35$ (Choice B) B $900\ln(1.25)-35$ (Choice C) C $900\ln(0.8)-175$ (Choice D) D $900\ln(1.25)-175$
Answer: Letting $C(x)$ be the consumer surplus when $x$ thousand units are sold, we are given that $C'(x)=\dfrac{900}{(x+13)}-35$. We want to find $C(12)-C(7)$. According to the Fundamental Theorem of Calculus, $\begin{aligned} C(12)-C(7)&=\int_{7}^{12} C'(x)\,dx \\\\ &=\int_{7}^{12}\left(\dfrac{900}{x+13}-35\right)dx \end{aligned}$ $\int_{7}^{12}\left(\dfrac{900}{x+13}-35\right)dx=900\ln(1.25)-175$ In conclusion, between $x=7$ and $x=12$ the consumer surplus increases by $900\ln(1.25)-175$ dollars.